How to determine Coercive functions












1












$begingroup$


A continuous function $f(x)$ that is defined on $R^n$ is called coercive if $limlimits_{Vert x Vert rightarrow infty} f(x)=+ infty$.



I am finding it difficult to understand how the norm of these functions are computed in order to show that they are coercive.

$a) f(x,y)=x^2+y^2
\b)f(x,y)=x^4+y^4-3xy\c)f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$



To show that they are coercive I have to show that as norm goes to infinity the function too should go to infinity right?






optimization nonlinear-optimization







share|cite|improve this question









$endgroup$












  • $begingroup$
    Yes, that is right.
    $endgroup$
    – uniquesolution
    Sep 8 '15 at 16:47










  • $begingroup$
    I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
    $endgroup$
    – MrYouMath
    Sep 8 '15 at 16:54










  • $begingroup$
    The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
    $endgroup$
    – K. Miller
    Sep 8 '15 at 17:06
















1












$begingroup$


A continuous function $f(x)$ that is defined on $R^n$ is called coercive if $limlimits_{Vert x Vert rightarrow infty} f(x)=+ infty$.



I am finding it difficult to understand how the norm of these functions are computed in order to show that they are coercive.

$a) f(x,y)=x^2+y^2
\b)f(x,y)=x^4+y^4-3xy\c)f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$



To show that they are coercive I have to show that as norm goes to infinity the function too should go to infinity right?






optimization nonlinear-optimization







share|cite|improve this question









$endgroup$












  • $begingroup$
    Yes, that is right.
    $endgroup$
    – uniquesolution
    Sep 8 '15 at 16:47










  • $begingroup$
    I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
    $endgroup$
    – MrYouMath
    Sep 8 '15 at 16:54










  • $begingroup$
    The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
    $endgroup$
    – K. Miller
    Sep 8 '15 at 17:06














1












1








1





$begingroup$


A continuous function $f(x)$ that is defined on $R^n$ is called coercive if $limlimits_{Vert x Vert rightarrow infty} f(x)=+ infty$.



I am finding it difficult to understand how the norm of these functions are computed in order to show that they are coercive.

$a) f(x,y)=x^2+y^2
\b)f(x,y)=x^4+y^4-3xy\c)f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$



To show that they are coercive I have to show that as norm goes to infinity the function too should go to infinity right?






optimization nonlinear-optimization







share|cite|improve this question









$endgroup$




A continuous function $f(x)$ that is defined on $R^n$ is called coercive if $limlimits_{Vert x Vert rightarrow infty} f(x)=+ infty$.



I am finding it difficult to understand how the norm of these functions are computed in order to show that they are coercive.

$a) f(x,y)=x^2+y^2
\b)f(x,y)=x^4+y^4-3xy\c)f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$



To show that they are coercive I have to show that as norm goes to infinity the function too should go to infinity right?






optimization nonlinear-optimization




optimization nonlinear-optimization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 8 '15 at 16:43









clarksonclarkson

87611535




87611535












  • $begingroup$
    Yes, that is right.
    $endgroup$
    – uniquesolution
    Sep 8 '15 at 16:47










  • $begingroup$
    I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
    $endgroup$
    – MrYouMath
    Sep 8 '15 at 16:54










  • $begingroup$
    The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
    $endgroup$
    – K. Miller
    Sep 8 '15 at 17:06


















  • $begingroup$
    Yes, that is right.
    $endgroup$
    – uniquesolution
    Sep 8 '15 at 16:47










  • $begingroup$
    I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
    $endgroup$
    – MrYouMath
    Sep 8 '15 at 16:54










  • $begingroup$
    The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
    $endgroup$
    – K. Miller
    Sep 8 '15 at 17:06
















$begingroup$
Yes, that is right.
$endgroup$
– uniquesolution
Sep 8 '15 at 16:47




$begingroup$
Yes, that is right.
$endgroup$
– uniquesolution
Sep 8 '15 at 16:47












$begingroup$
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
$endgroup$
– MrYouMath
Sep 8 '15 at 16:54




$begingroup$
I think your definition of coercive is not right. en.wikipedia.org/wiki/Coercive_function
$endgroup$
– MrYouMath
Sep 8 '15 at 16:54












$begingroup$
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
$endgroup$
– K. Miller
Sep 8 '15 at 17:06




$begingroup$
The definition of coercive is fine. It is the definition commonly used in nonlinear programming.
$endgroup$
– K. Miller
Sep 8 '15 at 17:06










2 Answers
2






active

oldest

votes


















0












$begingroup$

Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbf{x}) = |mathbf{x}|^2$. Now you can see that $f(mathbf{x}) to infty$ as $|mathbf{x}| to infty$.



Here is a hint for the second function. Use the inequality $-frac{3}{2}(x^2 + y^2) leq -3xy$ to derive a lower bound for $f(x,y)$. Show that for any $M > 0$ there exists a number $K > 0$ such that $f(x,y) > M$ whenever $sqrt{x^2 + y^2} > K$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    What I don't understand is what is this $||x||$
    $endgroup$
    – sam_rox
    Sep 8 '15 at 17:14












  • $begingroup$
    The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
    $endgroup$
    – K. Miller
    Sep 8 '15 at 17:20












  • $begingroup$
    @K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
    $endgroup$
    – clarkson
    Sep 8 '15 at 17:30










  • $begingroup$
    No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
    $endgroup$
    – K. Miller
    Sep 8 '15 at 17:35










  • $begingroup$
    @K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrt{x^8+y^8}$
    $endgroup$
    – clarkson
    Sep 8 '15 at 17:53



















0












$begingroup$

For (c),
use
$e^x ge 1+x$,
so
$e^{x^2} ge 1+x^2$.






share|cite|improve this answer









$endgroup$














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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbf{x}) = |mathbf{x}|^2$. Now you can see that $f(mathbf{x}) to infty$ as $|mathbf{x}| to infty$.



    Here is a hint for the second function. Use the inequality $-frac{3}{2}(x^2 + y^2) leq -3xy$ to derive a lower bound for $f(x,y)$. Show that for any $M > 0$ there exists a number $K > 0$ such that $f(x,y) > M$ whenever $sqrt{x^2 + y^2} > K$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      What I don't understand is what is this $||x||$
      $endgroup$
      – sam_rox
      Sep 8 '15 at 17:14












    • $begingroup$
      The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
      $endgroup$
      – K. Miller
      Sep 8 '15 at 17:20












    • $begingroup$
      @K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
      $endgroup$
      – clarkson
      Sep 8 '15 at 17:30










    • $begingroup$
      No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
      $endgroup$
      – K. Miller
      Sep 8 '15 at 17:35










    • $begingroup$
      @K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrt{x^8+y^8}$
      $endgroup$
      – clarkson
      Sep 8 '15 at 17:53
















    0












    $begingroup$

    Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbf{x}) = |mathbf{x}|^2$. Now you can see that $f(mathbf{x}) to infty$ as $|mathbf{x}| to infty$.



    Here is a hint for the second function. Use the inequality $-frac{3}{2}(x^2 + y^2) leq -3xy$ to derive a lower bound for $f(x,y)$. Show that for any $M > 0$ there exists a number $K > 0$ such that $f(x,y) > M$ whenever $sqrt{x^2 + y^2} > K$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      What I don't understand is what is this $||x||$
      $endgroup$
      – sam_rox
      Sep 8 '15 at 17:14












    • $begingroup$
      The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
      $endgroup$
      – K. Miller
      Sep 8 '15 at 17:20












    • $begingroup$
      @K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
      $endgroup$
      – clarkson
      Sep 8 '15 at 17:30










    • $begingroup$
      No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
      $endgroup$
      – K. Miller
      Sep 8 '15 at 17:35










    • $begingroup$
      @K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrt{x^8+y^8}$
      $endgroup$
      – clarkson
      Sep 8 '15 at 17:53














    0












    0








    0





    $begingroup$

    Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbf{x}) = |mathbf{x}|^2$. Now you can see that $f(mathbf{x}) to infty$ as $|mathbf{x}| to infty$.



    Here is a hint for the second function. Use the inequality $-frac{3}{2}(x^2 + y^2) leq -3xy$ to derive a lower bound for $f(x,y)$. Show that for any $M > 0$ there exists a number $K > 0$ such that $f(x,y) > M$ whenever $sqrt{x^2 + y^2} > K$.






    share|cite|improve this answer











    $endgroup$



    Consider the first function $f(x,y) = x^2 + y^2$. This function can be written in terms of vectors as $f(mathbf{x}) = |mathbf{x}|^2$. Now you can see that $f(mathbf{x}) to infty$ as $|mathbf{x}| to infty$.



    Here is a hint for the second function. Use the inequality $-frac{3}{2}(x^2 + y^2) leq -3xy$ to derive a lower bound for $f(x,y)$. Show that for any $M > 0$ there exists a number $K > 0$ such that $f(x,y) > M$ whenever $sqrt{x^2 + y^2} > K$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Sep 8 '15 at 18:49

























    answered Sep 8 '15 at 17:10









    K. MillerK. Miller

    3,673612




    3,673612












    • $begingroup$
      What I don't understand is what is this $||x||$
      $endgroup$
      – sam_rox
      Sep 8 '15 at 17:14












    • $begingroup$
      The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
      $endgroup$
      – K. Miller
      Sep 8 '15 at 17:20












    • $begingroup$
      @K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
      $endgroup$
      – clarkson
      Sep 8 '15 at 17:30










    • $begingroup$
      No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
      $endgroup$
      – K. Miller
      Sep 8 '15 at 17:35










    • $begingroup$
      @K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrt{x^8+y^8}$
      $endgroup$
      – clarkson
      Sep 8 '15 at 17:53


















    • $begingroup$
      What I don't understand is what is this $||x||$
      $endgroup$
      – sam_rox
      Sep 8 '15 at 17:14












    • $begingroup$
      The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
      $endgroup$
      – K. Miller
      Sep 8 '15 at 17:20












    • $begingroup$
      @K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
      $endgroup$
      – clarkson
      Sep 8 '15 at 17:30










    • $begingroup$
      No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
      $endgroup$
      – K. Miller
      Sep 8 '15 at 17:35










    • $begingroup$
      @K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrt{x^8+y^8}$
      $endgroup$
      – clarkson
      Sep 8 '15 at 17:53
















    $begingroup$
    What I don't understand is what is this $||x||$
    $endgroup$
    – sam_rox
    Sep 8 '15 at 17:14






    $begingroup$
    What I don't understand is what is this $||x||$
    $endgroup$
    – sam_rox
    Sep 8 '15 at 17:14














    $begingroup$
    The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
    $endgroup$
    – K. Miller
    Sep 8 '15 at 17:20






    $begingroup$
    The function $|cdot|$ is called a norm. In this case you are using the Euclidean norm or $2$-norm. For any vector $mathbf{x} = (x_1,ldots,x_n) in mathbb{R}^n$ the $2$-norm of $mathbf{x}$ is defined by $|mathbf{x}| = (x_1^2 + cdots + x_n^2)^{1/2}$. In two variables $x$ and $y$ you have $|(x,y)| = sqrt{x^2 + y^2}$.
    $endgroup$
    – K. Miller
    Sep 8 '15 at 17:20














    $begingroup$
    @K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
    $endgroup$
    – clarkson
    Sep 8 '15 at 17:30




    $begingroup$
    @K.Miller SO even for $f(x,y)=x^4+y^4-3xy$ as there are only two variables is the norm = $|(x,y)| = sqrt{x^2 + y^2}$. For $f(x,y,z)=e^{x^2}+e^{y^2}+e^{z^2}$ as there are three variables is the norm = $sqrt{x^2+y^2+z^2}$
    $endgroup$
    – clarkson
    Sep 8 '15 at 17:30












    $begingroup$
    No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
    $endgroup$
    – K. Miller
    Sep 8 '15 at 17:35




    $begingroup$
    No. The norm is just a function that maps vectors to a positive real number. You need to rearrange $f(x,y)$ to show that as $x^2 + y^2$ grows large so to does $f(x,y)$.
    $endgroup$
    – K. Miller
    Sep 8 '15 at 17:35












    $begingroup$
    @K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrt{x^8+y^8}$
    $endgroup$
    – clarkson
    Sep 8 '15 at 17:53




    $begingroup$
    @K.Miller I understand that I have to show function grows as norm grows. But my problem is finding out what is norm. Is the norm for $x^4+y^4-3xy$ is $sqrt{x^8+y^8}$
    $endgroup$
    – clarkson
    Sep 8 '15 at 17:53











    0












    $begingroup$

    For (c),
    use
    $e^x ge 1+x$,
    so
    $e^{x^2} ge 1+x^2$.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      For (c),
      use
      $e^x ge 1+x$,
      so
      $e^{x^2} ge 1+x^2$.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        For (c),
        use
        $e^x ge 1+x$,
        so
        $e^{x^2} ge 1+x^2$.






        share|cite|improve this answer









        $endgroup$



        For (c),
        use
        $e^x ge 1+x$,
        so
        $e^{x^2} ge 1+x^2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 8 '15 at 18:52









        marty cohenmarty cohen

        75.6k549130




        75.6k549130






























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            Wiesbaden

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            Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is...
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            Marschland

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            Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si...
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            Dieringhausen

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            Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7 ...
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